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Field if I define The plane

by Bachelier
Tags: define, field, plane
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Bachelier
#1
Jan30-13, 10:14 PM
P: 376
if I define The plane: ##F = ℝ## x ##ℝ = \{ (a, b) | a, b ∈ ℝ \} ##

and define addition and multiplication as:
(a, b) + (c, d) := (a + c, b + d)
(a, b) (c, d) := (ac, bd)

Then ##F## is a field. right?

would the multiplication as described here make ℂ a field?
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pwsnafu
#2
Jan30-13, 11:54 PM
Sci Advisor
P: 820
Quote Quote by Bachelier View Post
if I define The plane: ##F = ℝ## x ##ℝ = \{ (a, b) | a, b ∈ ℝ \} ##

and define addition and multiplication as:
(a, b) + (c, d) := (a + c, b + d)
(a, b) (c, d) := (ac, bd)

Then ##F## is a field. right?
No, you have zero divisors.

would the multiplication as described here make ℂ a field?
How would you do i2 = -1?
Bachelier
#3
Jan31-13, 12:21 AM
P: 376
Quote Quote by pwsnafu View Post
No, you have zero divisors.
You mean like (0,7).(8,0)

HallsofIvy
#4
Jan31-13, 09:46 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,316
Field if I define The plane

Yes, neither (7, 0) nor (0, 8) is the additive identity but neither has a mulitplicative inverse.
Bachelier
#5
Jan31-13, 11:35 AM
P: 376
thanks. Basically ℂ will fail to be an integral domain in the first place under this operation.


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